Conditional Holder inequality with $p=1,\infty$.

Question

For $X,Y\geq 0$ and $p \in (1,\infty)$ we have the conditional Holder inequality $$ E[XY|\mathcal{G}] \leq (E[X^p|\mathcal{G}])^{1/p}(E[Y^q|\mathcal{G}])^{1/q} $$ via some trickery using the usual Holder inequality. Does this have an extension to $p=1, q=\infty$? Clearly I can put $$ E[XY|\mathcal{G}] \leq E[X|\mathcal{G}] \cdot||Y||_\infty $$ but this is unsatisfying because it involves $Y$ on the right instead of $E[Y|\mathcal{G}]$. I would like to put $$ E[XY|\mathcal{G}] \leq E[X|\mathcal{G}] \cdot||E[Y|\mathcal{G}]||_\infty $$ or perhaps $$ E[XY|\mathcal{G}] \leq E[X|\mathcal{G}] \cdot \mathrm{ess\,sup}_p\,E[Y^p|\mathcal{G}]^{1/p}. $$